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Engineering 43. Sinusoidal AC SteadySt. Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu. Outline – AC Steady State. SINUSOIDS Review basic facts about sinusoidal signals SINUSOIDAL AND COMPLEX FORCING FUNCTIONS
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Engineering 43 SinusoidalAC SteadySt Bruce Mayer, PE Licensed Electrical & Mechanical EngineerBMayer@ChabotCollege.edu
Outline – AC Steady State • SINUSOIDS • Review basic facts about sinusoidal signals • SINUSOIDAL AND COMPLEX FORCING FUNCTIONS • Behavior of circuits with sinusoidal independent current & voltage sources • Modeling of sinusoids in terms of complex exponentials
phasEr Outline – AC SS cont. • PHASORS • Representation of complex exponentials as vectors • Facilitates steady-state analysis of circuits • Has NOTHING to do with StarTrek • IMPEDANCE AND ADMITANCE • Generalization of the familiar concepts of RESISTANCE and CONDUCTANCE to describe AC steady-state circuit operation
Outline – AC SS cont.2 • PHASOR DIAGRAMS • Representation of AC voltages and currents as COMPLEX VECTORS • BASIC AC ANALYSIS USING KIRCHHOFF’S LAWS • ANALYSIS TECHNIQUES • Extension of node, loop, SuperPosition, Thevenin and other KVL/KCL Linear-Circuit Analysis techniques
Recall From Trig the Sine Function Sinusoids • Where • XM “Amplitude” or Peak or Maximum Value • Typical Units = A or V • ω Radian, or Angular, Frequency in rads/sec • ωt Sinusoid argument in radians (a pure no.) • The function Repeats every 2π; mathematically • For the RADIAN Plot Above, The Functional Relationship
Now Define the “Period”, T Such That Sinusoids cont. • From Above Observe • How often does the Cycle Repeat? • Define Next the CYCLIC FREQUENCY • Now Can Construct a DIMENSIONAL (time) Plot for the sine
Now Define the Cyclic “Frequency”, f Sinusoids cont.2 • Quick Example • USA Residential Electrical Power Delivered as a 115Vrms, 60Hz, AC sine wave • Describes the Signal Repetition-Rate in Units of Cycles-Per-Second, or HERTZ (Hz) • Hz is a Derived SI Unit • Will Figure Out the 2 term Shortly • RMS “Root (of the) Mean Square”
Now Consider the GENERAL Expression for a Sinusoid Sinusoids cont.3 • Where • θ “Phase Angle” in Radians • Graphically, for POSITIVE θ
Consider Two Sinusoids with the SAME Angular Frequency Leading, Lagging, In-Phase • If = , Then The Signals are IN-PHASE • If , Then The Signals are OUT-of-PHASE • Phase Angle Typically Stated in DEGREES, but Radians are acceptable • Both These Forms OK • Now if > • x1 LEADS x2 by (−) rads or Degrees • x2 LAGS x1 by ( −) rads or Degrees
To Convert sin↔cos Useful Trig Identities • Additional Relations • To Make a Valid Phase-Angle Difference Measurement BOTH Sinusoids MUST have the SAME Frequency & Trig-Fcn (sin OR cos) • Useful Phase-Difference ID’s
Given Signals Example Phase Angles • To find phase angle must express BOTH sinusoids using • The SAME trigonometric function; • either sine or cosine • A POSITIVE amplitude • Find • Frequency in Standard Units of Hz • Phase Difference • Frequency in radians per second is the PreFactor for the time variable • Thus
Convert –6V Amplitude to Positive Value Using Example – Phase Angles cont. • It’s Poor Form to Express phase shifts in Angles >180° in Absolute Value • Then • Next Convert cosine to sine using • So Finally • Thus v1 LEADS v2 by 120°
Consider Instantaneous Power For a PurelyResistive Load Effective or rms Values • Since a Resistive Load Dissipates this Power as HEAT, the Effective Value is also called the HEATING Value for the Time-Variable Source • For example • A Car Coffee Maker Runs off 12 Vdc, and Heats the Water in 223s. • Connect a SawTooth Source to the coffee Maker and Adjust the Amplitude for the Same Time → Effective Voltage of 12V • Now Define The EFFECTIVE Value For a Time-Varying Signal as the EQUIVALENT DC value That Supplies The SAME AVERAGE POWER
For The Resistive Case, Define Ieff for the Avg Power Condition rms Values Cont. • If the Current is DC, then i(t) = Idc, so • The PavCalc For a Periodic Signal by Integ • Now for the Time-Variable Current i(t) → Ieff, and, by Definition
In the Pwr Eqn rms Values cont.2 • Examine the Eqn for Ieff and notice it is Determined by • Taking the Square ROOT of the time-averaged, or MEAN, SQUARE of the Current • In Engineering This Operation is given the Short-hand notation of “rms” • So • Equating the 1st & 3rd Expression for Pav find • This Expression Holds for ANY Periodic Signal
RMS Value for Sinusoid • Find RMS Value for sinusoidal current • Note the Period • Sub Above into RMS Eqn: • Use Trig ID:
RMS Value for Sinusoid • Sub cos2 Trig ID into RMS integral • Integrating Term by Term
RMS Value for Sinusoid • Rearranging a bit • But the integral of a sinusoid over ONE PERIOD is ZERO, so the 2nd Term goes to Zero leaving
For a Sinusoidal Source Driving a Complex (Z = R + jX) Load Sinusoidal rmsAlternative • Similarly for the rms Current • If the Load is Purely Resistive • Now, By the “effective” Definition for a R Load
In General for a Sinusoidal Quantity Sinusoidal rms Values cont • Thus the Power to a Reactive Load Can be Calculated using These Quantities as Measured at the SOURCE • Using a True-rms DMM • The rms Voltage • The rms Current • Using an Oscilloscope and “Current Shunt” • The Phase Angle Difference • For the General, Complex-Load Case • By the rms Definitions
Given Voltage Waveform Find the rms Voltage Value T Example rms Voltage • During the 2s Rise Calc the slope • m = [4V/2s] = 2 V/s • Thus The Math Model for the First Complete Period • Find The Period • T = 4 s • Derive A Math Model for the Voltage WaveForm • Use the rms Integral
Calc the rms Voltage T 0 Example rms Voltage cont. • Numerically
Given Current Waveform Thru a 10ΩResistor, then Find the Average Power Example Average Power • The “squared” Version • Then the Power • Find The Period • T = 8 s • Apply The rms Eqns
Consider the Arbitrary LINEAR Ckt at Right. If the independent source is a sinusoid of constant frequency then for ANY variable in the LINEAR circuit the STEADY-STATE Response will be SINUSOIDAL and of the SAME FREQUENCY Sinusoidal Forcing Functions • Thus to Find iss(t), Need ONLY to Determine Parameters A & • Mathematically
Given Simple Ckt Find i(t) in Steady State Write KVL for Single Loop Example RL Single Loop • In Steady State Expect • Sub Into ODE and Rearrange
Recall the Expanded ODE Example RL Single Loop cont • Equating the sin & cos PreFactors Yields • Solving for Constants A1 and A2 • Recognize as an ALGEBRAIC Relation for 2 Unkwns in 2 Eqns • Found A1 & A2using ONLY Algebra • This is Good
Using A1 & A2 State the i(t) Soln Example RL Loop cont.2 • Also the Source-V • If in the ID =x, and ωt = y, then • Would Like Soln in Form • Comparing Soln to Desired form → use sum-formula Trig ID
Dividing These Eqns Find Example RL Loop cont.3 • Now • Find A to be • Elegant Final Result, But VERY Tedious Calc for a SIMPLE Ckt • Not Good • Subbing for A & in Solution Eqn
Solving a Simple, One-Loop Circuit Can Be Very Tedious for Sinusoidal Excitations To make the analysis simpler relate sinusoidal signals to COMPLEX NUMBERS. The Analysis Of the Steady State Will Be Converted To Solving Systems Of Algebraic Equations ... Complex Exponential Form • Start with Euler’s Identity (Appendix A) • Where • Note: • The Euler Relation can Be Proved Using Taylor’s Series (Power Series) Expansion of ej
Now in the Euler Identity, Let Complex Exponential cont • So • Notice That if • Next Multiply by a Constant Amplitude, VM • Now Recall that LINEAR Circuits Obey SUPERPOSITION • Separate Function into Real and Imaginary Parts
Consider at Right The Linear Ckt with Two Driving Sources Complex Exponential cont.2 GeneralLinearCircuit • By KVL The Total V-src Applied to the Circuit • Now by SuperPosition The Current Response to the Applied Sources • This Suggests That the…
Application of the Complex SOURCE will Result in a Complex RESPONSE From Which The REAL (desired) Response Can be RECOVERED; That is Complex Exponential cont.3 GeneralLinearCircuit • Then The Desired Response can Be RECOVERED By Taking the REAL Part of the COMPLEX Responseat the end of theanalysis • Thus the To find the Response a COMPLEX Source Can Be applied.
We can NOT Build Physical (REAL) Sources that Include IMAGINARY Outputs Realizability GeneralLinearCircuit • We can also NOT invalidate Superposition if we multiply a REAL Source by ANY CONSTANT Including “j” • Thus Superposition Holds, mathematically, for • We CAN, However, BUILD These
This Time, Start with a COMPLEX forcing Function, and Recover the REAL Response at The End of the Analysis Let Example RL Single Loop • Thus Assume Current Response of the Form • In a Linear Ckt, No Circuit Element Can Change The Driving Frequency, but They May induce a Phase Shift Relative to the Driving Sinusoid • Then The KVL Eqn
Taking the 1st Time Derivative for the Assumed Solution Example – RL Single Loop cont. • Then the KVL Eqn • Then the Right-Hand-Side (RHS) of the KVL • Canceling ejt and Solving for IMej
Clear Denominator of The Imaginary Component By Multiplying by the Complex Conjugate Example – RL Single Loop cont.2 • Next A & θ • Then the Response in Rectangular Form: a+jb
First A Example – RL Single Loop cont.3 • A & Correspondence with Assumed Soln • A → IM • θ → • Cast Solution into Assumed Form • And θ
The Complex Exponential Soln Example – RL Single Loop cont.4 • Where • Recall Assumed Soln • Finally RECOVER the DESIRED Soln By Taking the REAL Part of the Response
By Superposition Example – RL Single Loop cont.5 • Explicitly • SAME as Before
If ALL dependent Quantities In a Circuit (ALL i’s & v’s) Have The SAME FREQUENCY, Then They differ only by Magnitude and Phase That is, With Reference to the Complex-Plane Diagram at Right, The dependent Variable Takes the form Imaginary A b a Real Phasor Notation • Borrowing Notation from Vector Mechanics The Frequency PreFactor Can Be Written in the Shorthand “Phasor” Form
Since in the Euler Reln The REAL part of the expression is a COSINE, Need to express any SINE Function as an Equivalent CoSine Turn into Cos Phasor Phasor Characteristics • Examples • Phasors Combine As the Complex Polar Exponentials that they are
As Before Sub a Complex Source for the Real Source Example RL Single Loop • The Form of the Responding Current • For the Complex Quantities • Phasor Variables Denoted as BOLDFACE CAPITAL Letters • Then The KVL Eqn in the Phasor Domain • Recall The KVL
To recover the desired Time Domain Solution Substitute Example RL Single Loop cont. • This is A LOT Easier Than Previous Methods • The Solution Process in the Frequency Domain Entailed Only Simple Algebraic Operations on the Phasors • Then by Superposition Take
The v-i Reln for R Resistors in Frequency Domain • Thus the Frequency Domain Relationship for Resistors
Phasors are complex numbers. The Resistor Model Has A Geometric Interpretation Resistors in -Land cont. • In the Complex-Plane The Current & Voltage Are CoLineal • i.e., Resistors induce NO Phase Shift Between the Source and the Response • Thus resistor voltage and current sinusoids are said to be “IN PHASE” R → IN Phase
The v-i Reln for L Inductors in Frequency Domain • Thus the Frequency Domain Relationship for Inductors
The relationship between phasors is algebraic. To Examine This Reln Note That Inductors in -Land cont. • Therefore the Current and Voltage are OUT of PHASE by 90° • Plotting the Current and Voltage Vectors in the Complex Plane • Thus
In the Time Domain Inductors in -Land cont.2 • Short Example L → current LAGS • Phase Relationship Descriptions • The VOLTAGE LEADS the current by 90° • The CURRENT LAGS the voltage by 90° • In the Time Domain